I can't find the $K_{sp}$ value for Manganese Phosphate ($\ce{Mn_3(PO_4)_2}$) anywhere online. Has it been measured? What is it?
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asked Apr 9, 2013 at 12:09
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Well, the MSDS entry tells me that the solubility is 50%. Converting, we get 0.5 g per gram of water, or 0.0005 g per liter of water. This is $s=1.4\times 10^{-6}$ moles per liter of water.
$K_{sp}=(3s)^3\times(2s)^2=6.14\times 10^{-27} \rm M^5 $
answered Apr 9, 2013 at 15:58
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$\begingroup$ We should get 0.5 g of manganese phosphate per gram of solution (the remaining 0.5 gram is water). Per liter, approximately 500 gram (.5 * 1000) is manganese phosphate and not 0.0005 g (.5 / 1000) - there are 1000 g of water in 1 L of water. $\endgroup$ Apr 9, 2013 at 18:07